Electrostatics 1 Question 15
16. Let $E_{1}(r), E_{2}(r)$ and $E_{3}(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. If $E_{1}\left(r_{0}\right)=E_{2}\left(r_{0}\right)$ $=E_{3}\left(r_{0}\right)$ at a given distance $r_{0}$, then
(2014 Adv.)
(a) $Q=4 \sigma \pi r_{0}^{2}$
(b) $r_{0}=\frac{\lambda}{2 \pi \sigma}$
(c) $E_{1}\left(r_{0} / 2\right)=2 E_{2}\left(r_{0} / 2\right)$
(d) $E_{2}\left(r_{0} / 2\right)=4 E_{3}\left(r_{0} / 2\right)$
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Answer:
Correct Answer: 16. (c)
Solution:
- $\frac{Q}{4 \pi \varepsilon_{0} r_{0}^{2}}=\frac{\lambda}{2 \pi \varepsilon_{0} r_{0}}=\frac{\sigma}{2 \varepsilon_{0}} \Rightarrow Q=2 \pi \sigma r_{0}^{2}$
(a) is incorrect, $r_{0}=\frac{\lambda}{\pi \sigma}$
(b) is incorrect, $E_{1} \frac{r_{0}}{2}=4 E_{1}\left(r_{0}\right)$
As $\quad E_{1} \propto \frac{1}{r^{2}}$
$$ E_{2} \frac{r_{0}}{2}=2 E_{2}\left(r_{0}\right) \text { as } E_{2} \propto \frac{1}{r} $$
(c) is correct
$$ E_{3} \frac{r_{0}}{2}=E_{3}\left(r_{0}\right)=E_{2}\left(r_{0}\right) $$
as $E_{3} \propto r^{0} \Rightarrow(\mathrm{d})$ option is incorrect
